# Dimension of a variety over finite field

Affine varieties, as the common zeroes of polynomials over some field, are irreducible algebraic sets and they're relatively easy to understand when polynomials are over $$\Bbb C$$. However, how can I determine that an algebraic set is a variety (i.e. irreducible) over a finite field $$\Bbb F_q$$, where q is a power of a prime. And more importantly, how to find its dimension?

For instance, in order to make everything easy, let $$f=x^2+3$$ and our finite field be $$\Bbb F_7$$. The algebraic set defined by $$f$$ over $$\Bbb F_7$$ is the set $$V= \{y\in acl(\Bbb F_7) : f(y)=0 \}$$

Therefore $$2,5,9,12 \in V$$, as $$2^2+3 \equiv5^2+3 \equiv 9^2+3 \equiv 0$$ (mod $$7$$), and there are infinitely such elements in $$V$$. Equivalently, $$V$$ consists of the numbers $$2$$ mod $$7$$ and $$5$$ mod $$7$$. So, I think that $$V$$ is defined by union of smaller algebraic sets $$x-2$$ and $$x-5$$. So, I think that V is not a variety. But, are these smaller components of $$V$$ varieties? If they're so, what are their dimension? More generally, how can I find dimension of a variety defined over a finite field. For example, what is dimension of the variety defined by $$x-2$$ over $$\Bbb F_7$$ ?

• 2,9 and 5,12 are the same elements modulo 7. – Wuestenfux Aug 14 '19 at 18:03
• @Wuestenfux I mean 9, 12 as elements of algebraic closure of F_7, which is union of all finite fields with 7^n elements for each n. So, 9 as itself stays in it, i guess? – offret Aug 14 '19 at 18:23
• In any field of characteristic $7$ you have $2=9$ and $5=12$. In particular for any algebraic closure of $\Bbb{F}_7$. – Servaes Aug 14 '19 at 19:21
• Chiming in with others. For a different way of looking at it: You do know that $\Bbb{F}_{7^n}\cong\Bbb{F}_7[x]/\langle p(x)\rangle$ for any irreducible polynomial $p(x)$ of degree from $\Bbb{F}_7[x]$. In the ring $\Bbb{F}_7[x]$ we have $2=9$, so the same holds for all the quotients. – Jyrki Lahtonen Aug 14 '19 at 19:50
• And yet another way of looking at it: The algebraic closure is an extension of $\Bbb{F}_7$. Meaning that the elements that are equal in $\Bbb{F}_7$ remain equal in all the extensions. For example $2/1$ and $4/2$ are the same elements in $\Bbb{Q}$, so they remain the same elements in $\Bbb{R}$ and $\Bbb{C}$. You need to review finite fields and field extension sooner rather than later. – Jyrki Lahtonen Aug 14 '19 at 19:58

Your $$V$$ consists of the union of two zero-dimensional varieties, namely the points $$y = 2$$ and $$y = -2$$. There are many ways to define the dimension of a variety, such as the Krull dimension (in the affine case) or the transcendence degree of the function field.